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A ladder of length $13 \mathrm{mts}$ has one end resting against a vertical wall and the other on the ground. If the lower end moves away from the wall at a speed of $2 \mathrm{mts} /$ minute, then the speed (in $\mathrm{mts} / \mathrm{min}$ ) at which upper end falls when the bottom is $5 \mathrm{mts}$ away from the wall is
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The correct answer is:
$\frac{5}{6}$
$\frac{d x}{d t}=2 \Rightarrow x^2+y^2=169$
$\begin{aligned} & 2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=0 \\ & \Rightarrow 2 x(2)+2 y\left(\frac{d y}{d t}\right)=0\end{aligned}$
when $x=5, y=12$
$\begin{aligned} & \Rightarrow 2(5)(2)+2(12) \frac{d y}{d t}=0 \\ & \Rightarrow \frac{d y}{d t}=\frac{-20}{24}=\frac{-5}{6}\end{aligned}$
$\therefore \quad$ Speed $=\frac{5}{6} \mathrm{~m} / \mathrm{min}$.
$\begin{aligned} & 2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=0 \\ & \Rightarrow 2 x(2)+2 y\left(\frac{d y}{d t}\right)=0\end{aligned}$
when $x=5, y=12$
$\begin{aligned} & \Rightarrow 2(5)(2)+2(12) \frac{d y}{d t}=0 \\ & \Rightarrow \frac{d y}{d t}=\frac{-20}{24}=\frac{-5}{6}\end{aligned}$
$\therefore \quad$ Speed $=\frac{5}{6} \mathrm{~m} / \mathrm{min}$.
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